N. Name: MATH: Mathematical Thinking Sec. 08 Spring 0 Worksheet 9: Solution Problem Compute the expected value of this probability distribution: x 3 8 0 3 P(x) 0. 0.0 0.3 0. Clearly, a value is missing from the table. In order to find P(x = 3) I should remember that the table describes a probability distribution. Then the sum should be equal to and So the expected value is P(x = 3) = (0. + 0.0 + 0.3 + 0.) = 0.8 = 0. E(x) = 3 0. + 8 0.0 + 0.3 + 0 0. + 3 0. = 0.6 + 0. +. + + 7 = 7. Problem Consider tossing times an unfair coin with a probability of 0.3 turning up heads. (a) Let x denote the number of heads in the tosses. Fill in the following probability distribution table for x. x 0 P(x) Using binomial probability it is easy to see that ( ) P(x = k) = (0.3) k (0.7) k k then the table is (b) Find the expected number of heads. x 0 P(x) 0.9 0. 0.09 Since we are using binomial probability E(x) = 0.3 = 0.6. That indeed is equal to E(x) = 0. + 0.09 = 0.6. Problem 3 An unfair coin with probability of heads 0. is tossed 3 times. Find the expected number of heads. Call x the number of heads in 3 tossing. As before, the probability distribution is given by the formula: ( ) 3 P(x = k) = (0.) k (0.8) 3 k k then the expectation value is: E(x) = 3 0. = 7 (using the expectation value for probability distribution). Problem Suppose you choose kids randomly from a group of girls and 8 boys. (a) Find the expected number of girls picked.
Call G the random variable that counts the number of girls.g can be {0,, }. ) and the expectation value is (b) Find the expected number of boys picked. P(G = 0) = P(G = ) = P(G = ) = 0( 0 ( 0 ) = 3 ) 3) = 8 ) 0 ) = E(G) = 0 3 + 8 + = 8 + = = 0.8 Call B the random variable that counts the number of boys. Then B = G and B can assume value {, 3, }. Obviously P(B = ) = P(G = 0) = 3 P(B = 3) = P(G = ) = 8 P(B = ) = P(G = ) = and the expectation value is E(G) = 3 + 3 8 + 0 + + = = 8 = 3. Of course, it s not an accident that E(G) + E(B) =. Problem Consider the following probability distribution of x, the number on a biased die. (a) Find P(x = 6). x 3 6 P(x) 0. 0. 0. 0. 0. In order to find P(x = 6) I should remember that the table describes a probability distribution. Then the sum should be equal to and (b) Find the expected value of x. The expected value is P(x = 6) = (0. + 0.0 + 0. + 0. + 0.) = 0.7 = 0.3 E(x) = 0. + 0. + 3 0. + 0. + 0. + 6 0.3 = 0. + 0. + 0.3 + 0.8 + 0. +.8 = 3.9 Problem 6 Let x be the number of a tiles drawn from a bag containing {,, 3,, }. (a) Fill in the following probability distribution for x. x 3 P(x) 0. 0. 0. 0. 0. since every event is equally likely so the probability is always the same, equal to (b) Find the expectation value of x.
The expectation value is: E(x) = 0. + 0. + 3 0. + 0. + 0. = ( + + 3 + + ) 0. = 0. = 3 Problem 7 (just for fun...) A raffle offers a first prize of 00$, 3 second prizes of 0$ and 0 third prizes of 60$. A total of $ tickets are sold. Determine the expected winnings for a person who buys ticket. (a) Let y denote the price of ticket. Determine the expected winnings for a person who buys ticket. Call x the money earned by a person who buys ticket. The probability distribution of x is x 399 9 9 -y P(x) 3 The last column is given by the fact that that person wins no prize, so there is a loss of y$. The expectation value is then E(x) = 399 (b) If y = 3$, is it worth it? In this case the expectation value is: 0 686 + 9 3 + 9 0 y 686 36 y 686 = E(x) = 36 3 686 = 6 0.88 Of course, the game is not fair, since on the average I will lose 88 cents, but this does not mean that it is not worth... (c) How much would you be willing to pay for a ticket? The best deal will be when E(x) = 0 This means 36 y 686 = 0 that is y = 36 686.0 Problem 8 In a game, you draw tiles (without replacement) from a bag that contains digit tiles: {,, 3,, }. If both of your tiles are even, you win $. If one of your tiles is even, you win 0.0$. Otherwise, you lose $. Find the expected gain or loss in one game. The random variable x can assume values {, 0., }. Call e the number of even tiles. P(x = ) = P(e = ) = P(x =.) = P(e = ) = P(x = ) = P(e = 0) = ( 0) ) = 0 ( ) ) = 6 0 0) ) = 3 0 Then the expectation value of x is: E(x) = 0. + 0. 0.6 0.3 = 0. + 0.3 0.3 = 0. Problem 9 I have two job interviews the same day. I have the 80% of chance of getting the first job, which pays 0,000$ per year. Instead, I have 0% of chance of getting the second job, which pays 30,000$ per year. (a) Which interview should I go to if I make my decision based on the expected gain of each interview? 3
This is the easiest case. Call X the possible gain if I go to the first job interview and X the possible gain if I go to the second job interview. Then, E(X ) = 0000 0.8 = 6000 E(X ) = 30000 0. = 000 and it is better if I go to the first job interview. (b) Suppose that I have a job that pays,000$ per year. Does my decision change? In this case, E(X ) = 0000 0.8 + 000 0. = 6000 + 3000 = 9000 E(X ) = 30000 0. + 000 0. = 000 + 700 = 00 So in this case it is better if I go to the second job interview. (c) Assume further that I have 0% of chance of getting a bonus of 0,000$ if I get the first job, and 30% of chance of getting a bonus of,000$ if I get the second job. Which will be my decision then? In order to see what it is better in this case, let s build up a diagram to summarize the situation. Call J, J the event that I got job or job, respectively, and B, B the event that I got the corresponding annual bonus. If I go to the first interview, then J 0.8 B X = 30000 0. 0.6 B X = 0000 0. J B X = 000 0 B X = 000 Of course, if I don t get the job it s impossible to get the bonus, so the probability that I gain,000$ per year is zero! The expectation value then is E(X ) = 30000 0. 0.8 + 0000 0.6 0.8 + 000 0 + 000 0. = 9600 + 9600 + 0 + 3000 = 00
If I go to the first interview, then B X = 000 0.3 J 0. 0.7 B X = 30000 0. J B X = 30000 0 B X = 000 The expectation value then is E(X ) = 000 0.3 0. + 30000 0.7 0. + 30000 0 + 000 0. = 670 + 000 + 0 + 700 = 0 So, even in this case it is better to take the risk.